Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 100010;
const int maxlog = 20;
int n, m, q;
vector<int> g[maxn];
int fa[maxn];
int L[maxn];
int anc[maxn][maxlog];
struct Node {
  int sz, a[15];
  Node() { sz = 0; }
  void add(int x) {
    for (int i = 0; i < sz; i++)
      if (a[i] == x) return;
    a[sz++] = x;
    sort(a, a + sz);
    if (sz > 10) sz--;
  }
} st[maxn][maxlog], val[maxn];
Node add(Node &x, Node &y) {
  Node rt = x;
  for (int i = 0; i < y.sz; i++) rt.add(y.a[i]);
  return rt;
}
void dfs(int u, int f, int d) {
  L[u] = d;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == f) continue;
    fa[v] = u;
    dfs(v, u, d + 1);
  }
}
void preprocess() {
  dfs(1, 0, 1);
  for (int i = 1; i <= n; i++) {
    anc[i][0] = fa[i];
    st[i][0] = add(val[i], val[fa[i]]);
    for (int j = 1; (1 << j) < n; j++) anc[i][j] = -1;
  }
  for (int j = 1; (1 << j) < n; j++)
    for (int i = 1; i <= n; i++)
      if (anc[i][j - 1] != -1) {
        int a = anc[i][j - 1];
        anc[i][j] = anc[a][j - 1];
        st[i][j] = add(st[i][j - 1], st[a][j - 1]);
      }
}
Node query(int p, int q) {
  Node ans;
  int tmp, log, i;
  if (L[p] < L[q]) swap(p, q);
  for (log = 1; (1 << log) <= L[p]; log++)
    ;
  log--;
  for (int i = log; i >= 0; i--)
    if (L[p] - (1 << i) >= L[q]) {
      ans = add(ans, st[p][i]);
      p = anc[p][i];
    }
  ans = add(ans, val[p]);
  if (p == q) return ans;
  for (int i = log; i >= 0; i--)
    if (anc[p][i] != -1 && anc[p][i] != anc[q][i]) {
      ans = add(ans, st[p][i]);
      ans = add(ans, st[q][i]);
      p = anc[p][i];
      q = anc[q][i];
    }
  ans = add(ans, st[p][0]);
  ans = add(ans, st[q][0]);
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int a;
    scanf("%d", &a);
    val[a].add(i);
  }
  preprocess();
  while (q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    Node ans = query(u, v);
    printf("%d", min(ans.sz, a));
    int cnt = 0;
    for (int i = 0; i < ans.sz; i++) {
      printf(" %d", ans.a[i]);
      if (++cnt == a) break;
    }
    puts("");
  }
}